The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned}\mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v)\rightleftharpoons & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(s o l v)+\mathrm{H}_{2} \mathrm{O}(so l v)\end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is \(6.68 .\) A pharmaceutical chemist makes up \(15.0 \mathrm{~L}\) of a solution that is initially \(0.275 \mathrm{M}\) in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Step by step solution

01

Write the reaction equation and define variables

The given reaction is: \(CH_3COOH(sol) + CH_3CH_2OH(sol) \rightleftharpoons CH_3COOCH_2CH_3(sol) + H_2O(sol)\) Let the initial moles of the reactants and products be: - \(I(CH_3COOH)= 0.275 \times 15.0 \,M\) - \(I(CH_3CH_2OH)= 3.85 \times 15.0 \,M\) - \(I(CH_3COOCH_2CH_3)= 0\,M \) (ethyl acetate is initially not present) - \(I(H_2O)= 0 \,M\) (initially not formed) Let x be the amount of acetic acid and ethanol reacted at equilibrium, then we have the change row as follows: Reaction: \(-x\,M \qquad -x \,M \qquad +x\,M \qquad +x\,M\)

02

Set up the ICE table

| | CH₃COOH | CH₃CH₂OH | CH₃COOCH₂CH₃ | H₂O | |---|--------|----------|-------------|-----| | I(initial) | 0.275M×15.0L | 3.85M×15.0L | 0 | 0 | | C(change) | -x | -x | +x | +x | | E(equilibrium)| 0.275M×15.0L-x|3.85M×15.0L-x| x | x |

03

Write the equilibrium expression and plug in values

The equilibrium constant expression for the reaction is: \[K_c = \frac{[CH_3COOCH_2CH_3][H_2O]}{[CH_3COOH][CH_3CH_2OH]}\] At equilibrium, we can substitute the values from the ICE table: \[6.68 = \frac{(x)(x)}{(0.275 \times 15.0 - x)(3.85 \times 15.0 - x)}\]

04

Solve for x

To solve for x, we can use the quadratic formula. However, as an approximation without losing much accuracy, we can consider the amount of reaction to be small compared to the initial concentrations, so: \(x \ll 0.275M \times 15.0L\) and \(x \ll 3.85M \times 15.0L\) Now, let's plug values and simplify: \(6.68 = \frac{x^2}{(0.275 \times 15.0)(3.85 \times 15.0)}\) \(x^2 = 6.68 \times (0.275 \times 15.0)(3.85 \times 15.0)\) \(x^2 = 529.31925\) Now, take the square root to find the value of x: \(x = \sqrt{529.31925} = 23.00998\, M\)

05

Convert equilibrium moles to grams of ethyl acetate

Now that we have the moles of ethyl acetate formed at equilibrium, we can convert it to grams using its molecular weight (\(CH_3COOCH_2CH_3\)), which is 88.11 g/mol: Grams of ethyl acetate formed at equilibrium = \(23.00998 \,M \times 88.11 \frac{g}{mol}\) Grams of ethyl acetate formed at equilibrium = \(2028.59002 \, g\) At equilibrium, approximately 2028.59 grams of ethyl acetate are formed.

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